1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

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1 1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the squres of the other two sides. Pythgors Theorem: + = How might one go out proving this is true? We n verify few exmples: Verify Pythgors theorem in the exmples elow In mthemtis this is not onsidered proof! Just euse this worked in these few exmples does not men tht it will lwys work. We need to give n rgument tht will work every time. The ide is to use geometry. Strt with generl right ngled tringle.

2 1 PYTHAGORAS THEOREM We re going to ssemle four opies of the right ngled tringle into squre, s follows. 1. Wht is the re of the red squre?. Wht is the re of the lue squre? 3. Wht is the re of one of the four tringles?

3 1 PYTHAGORAS THEOREM 3 Now lets put ll this together. Verify eh of the steps in the following lgeri rgument. re of red squre = re of lue squre + (4 re of tringle) ( ) 1 ( + ) = = + + = We n see now tht the geometri ft out res trnslted into Pythgors theorem! In this ws we n e sure tht no mtter whih right ngled tringle you give me, it will lwys e true tht + =.

4 WORKSHEET: EULER CHARACTERISTIC 4 Worksheet: Euler Chrteristi A grph is olletion of edges nd verties suh tht edges re not llowed to pss through one nother. Here re some exmples: Exmple 1 Exmple Exmple 3 We notie tht eh of the grphs stisfies the following properties: 1. Edges only meet t verties, tht is, edges do not ross one nother.. The grphs re ll onneted, tht is, ny vertex n e rehed from ny other y moving long pth of edges Here re some d exmples: Bd Grph 1 Bd Grph Question 1 Drw two exmples of grphs tht do stisfy our rules. Grph A Grph B A fe of grph is region ounded y edges. For instne, Exmple 1 hs 1 fe, Exmple hs 3 fes nd Exmple 3 hs 8 fes.

5 WORKSHEET: EULER CHARACTERISTIC 5 Question 1. Count how mny verties V, edges E nd fes F your two grphs hve. V E F Grph A Grph B. Now work out the vlue V E + F for oth of your exmples. Compre this vlue to those of your neighors. Wht do you notie? Grph A: V E + F = Grph B: V E + F = 3. Cn you mke guess t wht vlue V E + F might hve in generl? Question 3 Consider the following grph. We see tht V = 6 E = 6 F = 1 Now we re going to dd n edge to this grph V E + F = 1 This mens tht the numer of edges went up y 1, nd so did the numer of fes. Tht is V = 6 E = = 7 F = = V E + F = Now we repet this proess, dding one more edge nd fe eh time to ut out grph into tringles:

6 WORKSHEET: EULER CHARACTERISTIC 6 V E + F = V E + F = V E + F = 1. Wht do you notie out the vlue of V E + F eh time we dded tringle.. Cn you explin why this hppened? [Hint: Think out wht we re dding eh time, nd how tht effets eh quntity V, E nd F.] 3. Now we re going to remove tringles. V E + F = V E + F = V E + F = V E + F = Agin wht do you notie out the vlue V E + F, nd n you explin why this is the se?

7 3 WORKSHEET ON INDUCTION 7 3 Worksheet on Indution This worksheet we will e thinking out new kind of mthemtil proof, lled indution. As first exmple we ll thinking out wht hppens when we dd the first n numers, n It s often good ide to egin n investigtion suh s this y onsidering few smll exmples: n = 1 n = n = 3 1 = = = 6 We strt to notie pttern, nd mke guess tht n = 1. Verify this formul holds for the ses n = 1, nd 3 ove. n(n + 1) P(1) : 1 = 1 = 1 P() : 1 + = 3 P(3) : = 3 4 Does this formul hold for n = 100? How n we e sure? The nswer is mthemtil proof, of ourse! We need to prove tht this formul holds for ll n. We re going to strt the indution y proving tht P(4) holds without using the formul. Insted we will use P(3), whih we hve lredy verified to hold. P(4) : = ( ) + 4 ( ) 3 4 = + 4 (use P(3)) = (write 4 = 4 ) = = 4 5 (ommon denomintor). Prove tht the formul P(5) holds using only the formul P(4). [Hint: follow the sme reipe s ove in proving P(4)] 3. Prove tht the formul P(6) holds using only the formul P(5). Now lets try nd do something lever! We notie tht we n - use P(1) to prove P() - use P() to prove P(3) - use P(3) to prove P(4) - use P(4) to prove P(5)

8 3 WORKSHEET ON INDUCTION 8 - use P(5) to prove P(6) Cn we then use P(n 1) to prove P(n)? Let s try nd prove tht we n! We ssume tht we lredy know the sttement P(n 1). In other words this is the sttement tht the formul holds for numer n 1. Tht formul looks like this (n 1) = (n 1)((n 1) + 1) = (n 1)n Wht we wnt to do is use this formul to prove the next formul. Verify the following steps in this lgeri rgument (n 1) + n = ( (n 1)) + n ( ) (n 1)n = + n ( n ) n = + n = n n + n = n + = n(n + 1) Notie then tht we hve just shown tht the formul P(n) holds! And ll used ws the formul P(n 1). Another wy to sy this is tht if we know the formul holds for one numer, then we know it holds for the next numer. But we n repet this rgument gin nd gin. P(1) P() P(3) P(4) P(5) P(6) We n onlude then tht the formul P(n) holds for ny nturl numer n, nd this is n exmple of proof y indution!

9 4 ON THE SQUARE ROOT OF 9 4 On the squre root of A history of numers The first numers tht people disovered were the ounting numers 1,, 3, 4,... These re sometimes lled the nturl numers, euse they rise so nturlly ll round us. For exmple, you my hve 4 pples, nd use them to mke 1 pple pie, nd tht might mke 6 of your friends very hppy. After some time though people relized tht there were more numers lurking round. For exmple, when we finish the pie there will e 0 pie left (nd this mkes us sd). So we found the importnt numer zero. Soon fterwords negtive numers were hit upon (if we n go from 1 to 0, we should e le to go from 0 to 1, nd so on). So we eventully found the whole numers, whih we deided to ll the integers...., 3,, 1, 0, 1,, 3,... After little more pie mking ourred, we relised we needed even more numers! For exmple, wht hppens when we re hlf wy through the pie? We needed numers like 1 nd 3 4. We hd disovered the rtionl numers where nd re whole numers, or integers. The question now is, re we done? Do we hve ll the numers we ould ever need? Consider the unit squre. x 1 How long is the digonl line? We n nswer this question with Pythgors theorem. 1 x = = So x is the numer tht when squred gives us. We denote this numer y nd ll it the squre root of. Now is this numer one we hve lredy disovered? 1. Squre the following numers: () 1 = () = () 3 = (d) ( 1) = (e) ( ) =. Is nturl numer? 3. Is n integer? The only possiility we hve left is tht is rtionl numer. Tht is, there must exist integers nd suh tht =

10 4 ON THE SQUARE ROOT OF 10 Lowest form How mny different numers re there in the list elow: 1, 1, 4, 5 5, 1 3 We n represent rtionl numer uniquely y putting it into its lowest form, y whih we men, we nnot mke either or ny smller. Here re some exmples: = 8 = = 3 5 Question: Put the following rtionl numers into their lowest form We return to the numer. Suppose tht is rtionl. Then we n write it in its lowest form s = in lowest form If we multiply oth sides y, nd then squre oth sides, this eomes = whih we n interpret geometrilly s follows. The re of the the ig squre is twie the re of the smll squre. = + Exerise. Cut out opies of these squres (templtes found on the k pge). Now ple the red squre in the lower left orner of the lrger squre, nd the lue squre in the upper right orner. Notie tht you hve overed some prt of the lrge squre twie, nd some prts not t ll.

11 4 ON THE SQUARE ROOT OF 11 In prtiulr, notie tht sine the res of the two smller squres is supposed to omine to the re of the lrger squre, tht the prt we hve ounted twie (i.e. the red squre elow) hs the sme re s the sum of the two smller lue squres. = d + d d d

12 4 ON THE SQUARE ROOT OF 1 Question: 1. Wht re the sides lengths nd d of the red squre nd lue squre?. Are they integers? But now we hve prolem. Notie tht is smller thn, nd tht d is smller thn. Notie lso tht we hve the following eqution. = d whih if we tke squre roots nd divide y d eomes d = This mens tht = d ut sine nd d were smller this mens tht ws not in it s lowest form! This is ontrdition. Therefore we must hve mde flse ssumption. The only ssumption we mde ws tht ws rtionl numer, so this must in ft e flse! Let s summrize this disussion. Summry 1. Suppose tht is rtionl numer. Then we n write it in its lowest form = in lowest form. By onsidering this geometrilly we tully found smller numers, d tht were integers nd tht stisfied the sme ondition. = 3. This mens tht were not in lowest form. 4. This is ontrdition, so our originl ssumption must hve een flse! Therefore, d = not rtionl numer We hve therefore disovered new kind of numer. This new kind of numer is lled n irrtionl numer, euse it is not rtionl. Another exmple of n irrtionl numer is the fmous numer π tht reltes the rdius of irle to its irumferene.

13 4 ON THE SQUARE ROOT OF 13 Cutting Templtes

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